How many words can you make by permuting the letters of the word APPLE?

My attempt

$$\binom{5}{1}\binom{4}{2}\binom{2}{1}\binom{1}{1} = 60$$

What is the number of words you can make by permuting the letters of AAASB?

My attempt

$$\binom{5}{3}\binom{2}{1}\binom{1}{1} = 20$$

My question is why am I using the combination formula to find permutations? I'm having a mental block, or not seeing it clearly...... could someone clarify?

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    $\begingroup$ Do you want actual English words (or possibly other languages), or do you just want all permutations of the letters? $\endgroup$ – Dave Jul 5 '18 at 16:09
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    $\begingroup$ you can probably find your answer in this thread: math.stackexchange.com/questions/1424380/… $\endgroup$ – Rei Henigman Jul 5 '18 at 16:09
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    $\begingroup$ How many words of no greater than what length? $\endgroup$ – amWhy Jul 5 '18 at 16:09
  • $\begingroup$ just all permuations, im just confused why im using the combination formula $\endgroup$ – MasterYoshi Jul 5 '18 at 16:37
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    $\begingroup$ The nonsense word in the title doesnt match the text $\endgroup$ – David Diaz Jul 11 '18 at 12:52

For the number of permutations of $n$ letters, where one or more letter is repeated among the $n$ letters, we can use the multinomial theorem:

The multinomial coefficient is also the number of distinct ways to permute a multiset of $n$ elements, and $k_i$ are the multiplicities of each of the distinct elements.

For example, the number of distinct permutations of the letters of the word MISSISSIPPI, which has $1$ M, $4$ Is, $4$ Ss, and $2$ Ps and has $11$ characters all together (where $11$ is equal to the sum the multiplicities) is given by:

$$\binom{11}{1, 4, 4, 2} = \frac{11!}{1!\cdot 4! \cdot 4!\cdot 2!}$$

What we have in both cases is a word with five characters, each with characters that repeat.

The letters in APPLE have only the repetition of the letter P, and the other letters appear only once.. Then the number of permutations of APPLE is equal to $$\frac{5!}{1!1!1!2!} = \frac{5!}{2!} = 5\cdot 4\cdot 3 = 60 \text{ distinct permutations of the letters in APPLE}$$ where the denominator is the product of the number of occurrences of each letter in the original word (and they sum to $n = 5$).

In the case of AASSB, we have the letter A with two occurrences, the letter S with two occurrences, and the letter B with one occurrence:

The number of permutations of this number is given by $$\frac{5!}{2!2!1!} = \frac{5\cdot 4\cdot 3}{2} = 30 \text{ distinct permutations of the letters of AASSB}.$$

  • $\begingroup$ Wish you r my teacher amwhy. Good explanations step by step as always. Miss u. $\endgroup$ – mrs Jul 13 '18 at 14:13

By your approach, for the word APPLE, we are selecting

  • $1$ position among $5$ for the letter $A$ that is $\binom{5}{1}$
  • $2$ position among $4$ for the letters $P$ that is $\binom{4}{2}$
  • $1$ position among $2$ for the letter $L$ that is $\binom{2}{1}$
  • the position for the letter $E$ is then fixed

that is $$\binom{5}{1}\binom{4}{2}\binom{2}{1}=5\cdot 4\cdot 3=60$$

As an alternative, for APPLE we have $5!$ permutations for the $5$ letters but we need to divide by $2!$ because of the double $P$, indeed for example

$$\color{red}PAL\color{green}PE \quad \color{green}PAL\color{red}PE $$

are equivalent words and thus we obtain the same result

$$\frac{5!}{2!}=5\cdot 4\cdot 3=60$$

Can you proceed for $AASSB$?


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